Difference between revisions of "2009 AIME I Problems/Problem 2"

Revision as of 20:24, 2 September 2011

There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that

$\frac {z}{z + n} = 4i.$

Find $n$ .

1st Solution

Let $z = a + 164i$ .

Then $\frac {a + 164i}{a + 164i + n} = 4i$ and $a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.$

From this, we conclude that (HOW?) $a = -656$ and $164i = 4i \left (a + n \right ) = 4i \left (-656 + n \right ).$

We now have an equation for $n$ : $4i \left (-656 + n \right ) = 164i,$

and this equation shows that $n = \boxed{697}.$

2nd Solution

$\frac {z}{z+n}=4i$

$1-\frac {n}{z+n}=4i$

$1-4i=\frac {n}{z+n}$

$\frac {1}{1-4i}=\frac {z+n}{n}$

$\frac {1+4i}{17}=\frac {z}{n}+1$

Since their imaginary part has to be equal,

$\frac {4i}{17}=\frac {164i}{n}$

$n=\frac {(164)(17)}{4}=697$

$n = \boxed{697}.$

@@ Line 17: / Line 17: @@
 <cmath>a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.</cmath>
-From this, we conclude that
+From this, we conclude that (HOW?)
 <cmath>a = -656</cmath>
 and

2009 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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